I would rewrite the expression so that I could exclusively use the Product Rule and Chain Rule.
#f(x)=(x+13)^(1/2)(x-4)^(-1)(2x+1)^(1/3)#
#f'(x)=uvw'+uv'w+u'vw#
#u=(x+13)^(1/2)#
#u'=(1/2)(x+13)^(-1/2)#
#v=(x-4)^(-1)#
#v'=-(x-4)^-2#
#w=(2x+1)^(1/3)#
#w'=(1/3)(2x+1)^(-2/3)(2)#
#f'(x)=(x+13)^(1/2)(x-4)^(-1)(1/3)(2x+1)^(-2/3)(2)+(x+13)^(1/2)(-(x-4)^-2)(2x+1)^(1/3)+(1/2)(x+13)^(-1/2)(x-4)^(-1)(2x+1)^(1/3)#
Remove the negative exponents
#f'(x)=((x+13)^(1/2)(1/3)(2))/((x-4)(2x+1)^(2/3))+((x+13)^(1/2)(2x+1)^(1/3))/(-(x-4)^2)+((1/2)(2x+1)^(1/3))/((x+13)^(1/2)(x-4)^(1))#
Rearrange numerical constants and remove negatives from denominators
#f'(x)=(2(x+13)^(1/2))/(3(x-4)(2x+1)^(2/3))-((x+13)^(1/2)(2x+1)^(1/3))/((x-4)^2)+((2x+1)^(1/3))/(2(x+13)^(1/2)(x-4))#
Switch from exponents to radicals
#f'(x)=(2sqrt(x+13))/(3(x-4)root3((2x+1)^(2)))-(sqrt(x+13)root3(2x+1))/((x-4)^2)+(root3(2x+1))/(2sqrt(x+13)(x-4))#
